用于计算汇总ip地址的Bash脚本

问题描述

我需要编写一个bash脚本，当我输入两个ip地址时，它会为它们计算夏日地址。

I need to write a bash script that when i enter two ip addresses, it will calculate summerize address for them.

Examlpe：

192.168.1.27/25
192.168.1.129/25

结果将是：

192.168.1.0/24

你能帮我吗？用这个脚本？

Can you help me with this script?

我知道你会告诉我你尝试了什么。

I know you will say to me what did u try.

我试图在谷歌找到一些东西，但我发现我必须转换为二进制然后计算它，这将是非常困难的。

I tried to find something in Google, but what I found that I must to convert to binary then calculate it, and it will be very hard.

我甚至不知道如何开始：）

I even dont know how to start with it:)

有任何想法或提示吗？

谢谢，

Alaa

Thanks,
Alaa

推荐答案

常见的计算使用bash的网络掩码：

Calculation of common netmask with bash:

#!/bin/bash

D2B=({0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1})
declare -i c=0                              # set integer attribute

# read and convert IPs to binary
IFS=./ read -r -p "IP 1: " a1 a2 a3 a4     # e.g. 192.168.1.27
b1="${D2B[$a1]}${D2B[$a2]}${D2B[$a3]}${D2B[$a4]}"

IFS=./ read -r -p "IP 2: " a1 a2 a3 a4     # e.g. 192.168.1.129
b2="${D2B[$a1]}${D2B[$a2]}${D2B[$a3]}${D2B[$a4]}"

# find number of same bits ($c) in both IPs from left, use $c as counter
for ((i=0;i<32;i++)); do
  [[ ${b1:$i:1} == "${b2:$i:1}" ]] && c=c+1 || break
done

# create string with zeros
for ((i=c;i<32;i++)); do
  fill="${fill}0"
done

# append string with zeros to string with identical bits to fill 32 bit again
new="${b1:0:$c}${fill}"

# convert binary $new to decimal IP with netmask
new="$((2#${new:0:8})).$((2#${new:8:8})).$((2#${new:16:8})).$((2#${new:24:8}))/$c"
echo "$new"

输出：

192.168.1.0/24

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